SATHEE: Quantitative Aptitude Ques 730

Quantitative Aptitude Ques 730

Question: Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer.

I. $ \frac{18}{x^{2}}+\frac{6}{x}-\frac{12}{x^{2}}=\frac{8}{x^{2}} $ II. $ y^{3}+9.68+5.64=16.95 $

Options:

A) If $ x>y $

B) If $ x\ge y $

C) If $ x<y $

D) If $ x\le y $

E) If $ x=y $ or the relationship cannot be established

Show Answer

Answer:

Correct Answer: C

Solution:

I. $ \frac{18}{x^{2}}+\frac{6}{x}-\frac{12}{x^{2}}=\frac{8}{x^{2}} $
$ \Rightarrow $ $ \frac{6}{x^{2}}+\frac{6}{x}=\frac{8}{x^{2}} $

$ \Rightarrow $ $ \frac{6}{x}=\frac{2}{x^{2}} $
$ \Rightarrow $ $ 6x^{2}=2x $

$ \Rightarrow $ $ 2x(1-3x)=0 $
$ \Rightarrow $ $ x=0, $ $ \frac{1}{3} $ II. $ y^{3}=16.95-(9.68+5.64=16.95-15.32 $

$ \Rightarrow $ $ y^{3}=1.63 $
$ \Rightarrow $ $ y=1.176 $

$ \therefore $ $ y>x $

Quantitative Aptitude Ques 729

Quantitative Aptitude Ques 731

Bank Preparation

IBPS Office Assistants

IBPS RRB Officers

Bank Courses

Recorded Course for English

Recorded Course for Reasoning

Recorded Course for Mathematics

NCERT Books

Important Resources

Banking Awareness

Financial Awareness

Forum

BETA

© 2025 Copyright SATHEE

Privacy Policy

Powered by Prutor@IITK

sathee@iitk.ac.in

Media coverage of SATHEE

Play Store

App Store

The Ministry of Education has launched the SATHEE initiative in association with IIT Kanpur to provide free guidance for competitive exams. SATHEE offers a range of resources, including reference video lectures, mock tests, and other resources to support your preparation. Please note that participation in the SATHEE program does not guarantee clearing any exam or admission to any institute.